这里有一个简单的解决方案，以确保你的应用程序可以访问互联网:

static final String CHECK_INTERNET_ACCESS_URL = "https://www.google.com";

public static void isInternetAccessWorking(Context context) {

    StringRequest stringRequest = new StringRequest(Request.Method.GET, CHECK_INTERNET_ACCESS_URL,
            new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {
                    // Internet access is OK
                }
            }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            // NO internet access
        }
    });

    Volley.newRequestQueue(context).add(stringRequest);
}

这个解决方案使用Android的Volley库，必须在build.gradle中声明:

implementation 'com.android.volley:volley:1.1.1'

在芬兰湾的科特林:

class UtilityMethods {
companion object {
    fun isConnected(activity: Activity): Boolean {
        val connectivityManager: ConnectivityManager =
                activity.getSystemService(Context.CONNECTIVITY_SERVICE) as ConnectivityManager
        return null != connectivityManager.activeNetworkInfo
    }
}}

在Activity类中调用isConnected如下:

UtilityMethods.isConnected(this)

内部片段类如下:

UtilityMethods.isConnected(activity)

最好的方法:

public static boolean isOnline() {
    try {
    InetAddress.getByName("google.com").isReachable(3);

    return true;
    } catch (UnknownHostException e){
    return false;
    } catch (IOException e){
    return false;
    }
    }

检查这段代码…这对我很有用:)

public static void isNetworkAvailable(final Handler handler, final int timeout) {
    // ask fo message '0' (not connected) or '1' (connected) on 'handler'
    // the answer must be send before before within the 'timeout' (in milliseconds)

    new Thread() {
        private boolean responded = false;   
        @Override
        public void run() { 
            // set 'responded' to TRUE if is able to connect with google mobile (responds fast) 
            new Thread() {      
                @Override
                public void run() {
                    HttpGet requestForTest = new HttpGet("http://m.google.com");
                    try {
                        new DefaultHttpClient().execute(requestForTest); // can last...
                        responded = true;
                    } 
                    catch (Exception e) {
                    }
                } 
            }.start();

            try {
                int waited = 0;
                while(!responded && (waited < timeout)) {
                    sleep(100);
                    if(!responded ) { 
                        waited += 100;
                    }
                }
            } 
            catch(InterruptedException e) {} // do nothing 
            finally { 
                if (!responded) { handler.sendEmptyMessage(0); } 
                else { handler.sendEmptyMessage(1); }
            }
        }
    }.start();
}

然后，我定义处理程序:

Handler h = new Handler() {
    @Override
    public void handleMessage(Message msg) {

        if (msg.what != 1) { // code if not connected

        } else { // code if connected

        }   
    }
};

.．.并启动测试:

isNetworkAvailable(h,2000); // get the answser within 2000 ms

不需要太复杂。最简单的框架方式是使用ACCESS_NETWORK_STATE权限并创建一个连接的方法

public boolean isOnline() {
    ConnectivityManager cm =
        (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);

    return cm.getActiveNetworkInfo() != null && 
       cm.getActiveNetworkInfo().isConnectedOrConnecting();
}

如果您有特定的主机和连接类型(wifi/移动)，也可以使用requestRouteToHost。

你还需要:

<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />

在你的android清单中。

最简单的解决办法是

在大多数情况下，如果他/她想连接到远程服务器，只会检查互联网连接，所以简单和最好的解决方案是ping你的服务器，如下所示。

public boolean isConnected() {
    final String command = "ping -c 1 yourExmapleDomain.com";
    boolean isConnected = false;
    try {
        isConnected = Runtime.getRuntime().exec(command).waitFor() == 0;
    } catch (InterruptedException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    return isConnected;
}